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An infinitesimal coordinate change generated by $\xi$ changes a metric perturbation about Minkowski spacetime by $h_{\mu \nu }\mapsto h_{\mu \nu }-{\partial }_{\mu }{\xi }_{\nu }-{\partial }_{\nu }{\xi }_{\mu }$, in the passive sign convention. For a passive coordinate change $x^{\mu }\mapsto x^{\mu }+{\xi }^{\mu }$, the first-order change is minus the Lie derivative of a tensor field of the background metric, equivalently the pullback by the inverse flow. The gauge invariance of the linearized Riemann tensor follows because the added terms have commuting third derivatives, which cancel in the curvature antisymmetrizations. Gauge choices alter the metric components while preserving linearized tidal effects.

Use $c=1$, background Minkowski spacetime metric tensor ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$, and $g_{ab}={\eta }_{ab}+h_{ab}$ with small metric perturbation. All indices and derivatives in this linear approximation use $\eta$. The linearized inverse metric is ${\eta }^{ab}-h^{ab}$, and the linearized Levi-Civita connection isDefine the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$.

An infinitesimal coordinate change $x^{\prime a}=x^a+{\xi }^a$ gives the linearized coordinate gauge transformationThe metric perturbation therefore has redundant components. The gauge invariance of the linearized Riemann tensor follows because its change contains third derivatives of $\xi$ cancelling by commutation of partial derivatives. Conversely, flat linearized metric perturbations are locally pure gauge on a contractible patch. Thus linearized curvature, rather than a particular coordinate value of $h$, detects the physical disturbance. Gauge freedom can also be restricted by global topology or boundary conditions.

Keeping the paper's curvature sign convention, its linearized Ricci tensor and scalar areThese are the negatives of the alternative frequently used curvature convention. Put $v_b={\partial }^a{\bar{h}}_{ab}$. The Linearized Einstein equations in vacuum areUnder a gauge change $v_b\mapsto v_b-\square {\xi }_b$, so solving $\square {\xi }_b=v_b$ imposes the Lorenz gauge in linearized gravity. The field equations then become $\square {\bar{h}}_{ab}=0$. The remaining residual gauge symmetry of linearized gravity has $\square {\xi }_b=0$; Lorenz gauge does not exhaust the coordinate freedom.

A Fourier mode ${\bar{h}}_{ab}=A_{ab}{e}^{i{k}_c{x}^c}$ obeys $k^a{k}_a=0$ and $k^a{A}_{ab}=0$. Hence the plane gravitational waves in linearized gravity propagate on the background light cones. Four transversality conditions and four residual gauge amplitudes leave $10-4-4=2$ physical degrees of freedom. The explicit plane-wave reduction to transverse-traceless gauge can set $h_{0a}=0$, spatial trace zero and $k^i{h}_{ij}=0$. For a wave in the $z$ direction, write its physical spatial strain as $H_{ij}=-h_{ij}$, so the spatial metric tensor is $-({\delta }_{ij}+H_{ij})$:There are two gravitational wave polarizations, plus polarization and cross polarization. Rotating transverse axes through $\theta$ rotates their amplitude pair through $2\theta$. Nonconstant profiles with nonzero second derivative have nonzero linearized Riemann curvature tensor and cannot be removed as pure gauge.

The TT coordinate separation and measured separation distinction explains a detector's response. Freely falling particles initially at rest can keep fixed TT coordinates, since ${\Gamma }^a{}_{00}=0$, while their proper separation changes. For a short arm along a unit vector $n$,This is the tidal response described by geodesic deviation; coordinate motion by itself is not the measured signal.

Finally, gravitational-wave energy is second order in the perturbation, so it is absent from a first-order vacuum equation. In a short-wavelength averaging regime the averaged stress-energy of transverse gravitational waves isFor a plane wave its energy flux is $(16\pi G{)}^{-1}⟨{\dot{H}}_{+}^2+{\dot{H}}_{\times }^2⟩$. The averaging scale and weak-field assumptions matter: this does not assign a coordinate-independent local gravitational energy density to an arbitrary first-order perturbation.

Raise perturbation indices with the background Minkowski metric. To first order, the inverse metric is $g^{ik}={\eta }^{ik}-ϵh^{ik}+O({ϵ}^2)$ and the Levi-Civita connection isQuadratic connection products in the Riemann curvature tensor are of order ${ϵ}^2$. Substitution into the paper's curvature convention givesThe signs follow the convention on the cover sheet; reversing the curvature definition would reverse this expression.

The gauge freedom is the freedom to identify points of the perturbed spacetime with points of the flat background in slightly different coordinates. Under the passive infinitesimal change $x^{\prime i}=x^i+ϵ{\xi }^i(x)$, the metric transformation law gives the linearized coordinate gauge transformationThus a nonzero perturbation can partly represent a coordinate change rather than physical curvature. Substituting this variation into the displayed curvature formula givesEach third derivative has a partner differing only by the order of its commuting partial derivatives. This explicitly proves gauge invariance of the linearized Riemann tensor. The restriction to first order matters: a coordinate change acts on nonzero background curvature as well, but the background here is flat.

Define the trace-reversed metric perturbation and d'Alembert operator byIn four dimensions $\bar{h}=-h$ and $h_{ik}={\bar{h}}_{ik}-\frac{1}{2}{\eta }_{ik}\bar{h}$. Contract the curvature formula using $R_{ik}=R^m{}_{imk}$:up to $O({ϵ}^2)$. Consequently the Einstein tensor isChoose Lorenz gauge in linearized gravity, ${\partial }^k{\bar{h}}_{ik}=0$. It can be imposed locally by solving $\square {\xi }_i={\partial }^k{\bar{h}}_{ik}$, since the gauge transformation changes this divergence by $-\square {\xi }_i$. The remaining coordinate freedom satisfies $\square {\xi }_i=0$. In this gauge,This is the requested Linearized Einstein equations in the stated sign convention. Their divergence requires ${\partial }^k{T}_{ik}=0$ to this order.

For the photon beam, introduce $u=t-x$ and the constant null covector ${\ell }_i=(1,-1,0,0)$, so ${\ell }^i=(1,1,0,0)$. The stress-energy tensor isIt is trace-free and conserved: ${\ell }^i{\partial }_i\rho (u)=0$. A beam-adapted particular solution, with no independently added homogeneous gravitational field, isThe equality follows because its trace is zero, and the Lorenz condition holds because ${\ell }^k{\partial }_{k}H=0$. The field equation becomes a transverse Poisson equation:For the transversely uniform source in the question, one particular solution is $H=4\pi G\rho (u)(y^2+z^2)$. Thus the sourced components can be chosen asBoundary conditions and homogeneous solutions determine additional freedom; the source does not uniquely specify every component in every gauge. In particular, taking $H$ to depend only on $t-x$ would give $\square H=0$ and cannot solve the equation for nonzero $\rho$. The uniform idealization is not an asymptotically flat finite beam, and its quadratic particular solution is a weak-field description only where $|ϵH|\ll 1$.

Finally consider another photon with tangent proportional to ${\ell }^i$. Its tangent remains null because $h_{ik}{\ell }^i{\ell }^k=0$. Moreover $h_{ik}{\ell }^k=0$, and since $\ell$ is constant, the connection contraction isThe geodesic equation is therefore satisfied by a straight affinely parametrized ray along the beam. Hence co-propagating photons experience no deflection from the beam's gravitational field. This is parallel photon beams in linearized gravity; it does not assert absence of forces on counter-propagating photons or effects from unrelated homogeneous gravitational perturbations.

Let $\delta h_{\mu \nu }={\partial }_{\mu }{V}_{\nu }+{\partial }_{\nu }{V}_{\mu }$. Substitution into the linearized Levi-Civita connection cancels four terms and leavesTo first order in the metric perturbation, the products of Christoffel symbols are quadratic and may be discarded. Therefore the change in the Riemann curvature tensor isThis is the gauge invariance of the linearized Riemann tensor. Lowering the first index with the Minkowski metric does not change the conclusion; the change from using the full metric tensor instead is of second order because the background curvature is zero. HenceThe result is invariance at linear order about flat spacetime, not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of $V$ cancel in pairs by commuting partial derivatives.

In the Levi-Civita connection, the background metric is constant and replacing $g^{\alpha \rho }$ by its $O(ϵ)$ correction would multiply a derivative of $h$ and produce an $O({ϵ}^2)$ term. Hence the linearized Levi-Civita connection isThe products of two connection coefficients in the Riemann curvature tensor are also quadratic and may be discarded. Lowering its first index with ${\eta }_{\mu \nu }$ gives

Because the coordinate change is ${\tilde{x}}^{\alpha }=x^{\alpha }-{\xi }^{\alpha }$, the perturbation changes at linear order bySubstitution into $R_{\mu \rho \alpha \beta }^{(1)}$ produces terms containing three partial derivatives of $\xi$. Since partial derivatives commute, every term cancels another with the opposite sign. ThereforeThis is the gauge invariance of the linearized Riemann tensor.